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The sum of the first six terms of an arithmetic progression is $42$. The ratio of the $10^{\text{th}}$ term to the $30^{\text{th}}$ term is $1 : 3$. Calculate the first and the thirteenth terms of the AP.
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Let first term = $a$ and common difference = $$\begin{aligned}& d \\ & text{ATQ, } \frac{a_{10}}{a_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3} \\ & 3a + 27d = a + 29d \\ & 2a = 2d \Rightarrow a = d \\ & S_6 = \frac{6}{2}[2a + (6-1)d] = 42 \\ & 3[2a + 5a] = 42 \quad (\text{since } a=d) \\ & 3[7a] = 42 \\ & 21a = 42 \\ & a = 2 \\ & d = 2 \\ & a_{13} = a + 12d = 2 + 12 \times 2 = 2 + 24 = 26\end{aligned}$$